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Math Help - Basic Probability Proof

  1. #1
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    Basic Probability Proof

    P(A|C)≥P(B|C) and P(A|C')≥P(B|C')

    Prove that P(A)≥P(B).

    First I started with,
    P(A|C)=P(AC)/P(C) and similarly P(B|C)=P(BC)/P(C).

    From above,
    P(AC)/P(C)≥P(BC)/P(C) and since P(C)≥0,

    P(AC)≥P(BC).

    Similarly from above,
    P(AC')≥P(BC').

    Also, P(AC)=P(A)+P(C)-P(AUC) which led to what is below.

    As of now I've gotten down to... but got stuck

    P(A)≥P(B)-P(BUC)+P(AUC)
    P(A)≥P(B)-P(BUC')+P(AUC')
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  2. #2
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    Re: Basic Probability Proof

    Quote Originally Posted by Hal2001 View Post
    P(A|C)≥P(B|C) and P(A|C')≥P(B|C')
    Prove that P(A)≥P(B).
    It is so easy.
    From the given we have: P(AC)>P(BC)~\&~P(AC')>P(BC').
    But simply add those two and realize that P(A)=P(AC)+P(AC').
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  3. #3
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    Re: Basic Probability Proof

    Got it. Hmm... I really was making it much harder than it should have been. Thanks.
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